1. Use the table to complete the statement.

Question

Desmos Calculator (click here)

x Substituted Evaluate True Statement?
−34

−12

12



(1 point)
For the quadratic equation −2x2−3x+2=0
, complete the table by testing each of the given values to determine whether it is a solution. Identify which one of the values is in the solution set. Only x =

is in the solution set.
Question 2
2. Solve x2=289
by inspection. There are two real solutions. Enter the lesser number first.

Desmos Calculator (click here)

(2 points)
{
,
}
Question 3
3. Which values are in the solution set of the quadratic equation, −9+(x−3)2−4=−4
? Solve using square roots and select all that apply.

Desmos Calculator (click here)

(2 points)
Responses

−2
negative 2

16
16

28
28

−10
negative 10

8
8

2
2

−8
negative 8

−22
negative 22
Skip to navigation

page 2 of 3

Answer 1

To solve the problems presented, let's break them down.

Question 1

We need to complete the table for the quadratic equation \(-2x^2 - 3x + 2 = 0\) by substituting given values of \(x\) (-34, -12, 12) into the equation and evaluating whether the expression equals zero.

1. For \(x = -34\): \[ -2(-34)^2 - 3(-34) + 2 = -2(1156) + 102 + 2 = -2312 + 102 + 2 = -2210 \quad (\text{not a solution}) \]

2. For \(x = -12\): \[ -2(-12)^2 - 3(-12) + 2 = -2(144) + 36 + 2 = -288 + 36 + 2 = -250 \quad (\text{not a solution}) \]

3. For \(x = 12\): \[ -2(12)^2 - 3(12) + 2 = -2(144) - 36 + 2 = -288 - 36 + 2 = -322 \quad (\text{not a solution}) \]

None of these values make the quadratic equation equal to zero, indicating that none of them are solutions.

However, as no values equate to zero, what needs to be completed is simply confirming that none of the values are in the solution set. With the question stating "Only x =", this may require clarification as no options were valid.

Question 2

To solve the equation \(x^2 = 289\) by inspection, we can take the square root of both sides:

\[ x = \pm \sqrt{289} \]

Calculating the square root:

\[ \sqrt{289} = 17 \]

Thus, the solutions are:

\[ x = 17 \quad \text{and} \quad x = -17 \]

The two solutions (lesser first) are:

\[ {-17, 17} \]

Question 3

To solve the equation \(-9 + (x - 3)^2 - 4 = -4\), we simplify it:

First, add 9 to both sides:

\[ (x - 3)^2 - 4 = 5 \]

Then add 4 to both sides:

\[ (x - 3)^2 = 9 \]

Next, take the square root of both sides:

\[ x - 3 = \pm 3 \]

From this, we find two equations:

1. \(x - 3 = 3 \Rightarrow x = 6\)
2. \(x - 3 = -3 \Rightarrow x = 0\)

Thus, the solutions to the equation are \(x = 6\) and \(x = 0\).

Summary of Responses:

1. Only x = __ has no solutions in the provided values.
2. Solutions for \(x^2 = 289\) are: \({-17, 17}\)
3. The values in the solution set for the quadratic equation are: \(0\) and \(6\).

Please verify if you need any further adjustments or clarifications!